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I read the whole proof of Theorem 13.18 Bruckner's Real Analysis book and I had no problem understanding the proof except for following two claims inside the proof that are stated without further explanations.

One is dense-ness of $L^{\infty}$ in $L^{p}$ which the post in here is not very clear. How $L^{\infty}$ is dense in $L^{p}$?

The second : because two functionals $\Gamma_1, \Gamma : L^{p} \to \mathbb{F}$ agree on a dense subspace of $L^{p}$ how that implies they agree on whole domain $L^{p}$?

I would appreciate a detailed guidance about the two questions.

  • I don't have the book on hand. Are the functionals bounded? – Michael Jesurum Jun 17 '21 at 14:56
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    @MichaelJesurum, yes they are. I will add the link for the book in OP. –  Jun 17 '21 at 15:02
  • @L.G. Simple functions belong to $L_\infty$. Integrable simple functions are dense in $L_p$, so $L_\infty$ is dense in $L_p$ for any $p>0$ – Mittens Jun 17 '21 at 17:16
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    As for your question of equality between bounded linear functionals that coincide in dense sets, in addition to one of the solutions to your OP, you may see the comment section of this posting. The in fact holds for Lipschitz functions too. – Mittens Jun 17 '21 at 17:20

2 Answers2

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Item 1: The Theorem 13.18 is about $\sigma$-finite measure spaces and $L^p$, where $1\leq p < \infty$. The proof begins by considering the case of finite measure spaces.

So, let $(X, \mathcal{M}, \mu)$ be a measure space and $\mu(X) <\infty$. Given any $g \in L^\infty$, we have that there is $M>0$ such that $|g| \leq M$ almost everywhere. So, given any $p$, $1\leq p < \infty$, we have: $$\int_X |g|^p d\mu \leq \int_X M^p d\mu = M^p \mu(X) < \infty$$ So, we have that $L^\infty \subseteq L^p$.

Now, given any $f \in L^p$, then, for all $n \in \Bbb N$, let $f_n= f \cdot \chi_{\{x \in X : |f(x)| \leq n\}}$. It is easy to see that $f_n \in L^\infty$.

Note that $| f - f_n|^p = |f|^p \cdot \chi_{\{x \in X : |f(x)| > n\}}$. Since $f \in L^p$, $f$ is finite almost everywhere, so, as $n \to \infty$, we have $|f|^p \cdot \chi_{\{x \in X : |f(x)| > n\}} \to 0$ almost everywhere. Note also that $|f|^p \cdot \chi_{\{x \in X : |f(x)| > n\}} \leq |f|^p$ and, since $f$ is in $L^p$,we have that $|f|^p \in L^1$. Then, applying the Dominated Convergence Theorem, we have $$ \lim_n \int_X | f - f_n|^p d\mu = \lim_n \int_X |f|^p \cdot \chi_{\{x \in X : |f(x)| > n\}} d\mu = \int_X 0 \ d\mu =0$$

Remark: After proving $L^\infty \subseteq L^p$, if you know that the simple functions (which are functions in $L^\infty$) are dense in $L^p$, you can use this fact to deduce that $L^\infty$ is dense in $L^p$.

Item 2: You ask: "because two functionals $\Gamma_1, \Gamma : L^{p} \to \mathbb{F}$ agree on a dense subspace of $L^{p}$ how that implies they agree on whole domain $L^{p}$?"

The answer is yes, because the functionals $\Gamma_1, \Gamma : L^{p} \to \mathbb{F}$ in Theorem 13.18 are continuous linear functionals.

In fact, as QuantumSpace noted, this is a general property:

Let $X$ and $Y$ be metric spaces and let $D$ be a dense subset of $X$. Then, for any two continuous functions $f: X \rightarrow Y$ and $g:X \rightarrow Y$, if $f$ and $g$ agrees on $D$ then $f=g$

Proof: Given any $x \in X$, let $\{x_n\}_n$ be a sequence of points in $D$ such that $x_n \to x$. Then $$ f(x) = \lim_n f(x_n) = \lim_n g(x_n) =g(x) $$

Ramiro
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    Why the downvote? The OP (student) have clearly asked about Theorem 13.18, about two claims inside the proof that are stated without further explanations. My answer explains the two claims and it is correct. – Ramiro Jun 17 '21 at 17:30
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    +1, Not just your answers, someone always waiting for me posting a question then downvotes it. Almost all questions that I've posted in the last month had at least one downvote! Another strange thing is that when you write MUCH better answer your votes are 1 or 2 and the user's imperfect answer gets 3 or 4 upvotes! –  Jun 17 '21 at 17:57
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    Good answer, IMO should be accepted. The one thing I think would be nice is in the last part where you said, "In fact, this is a general property", if you said something along the lines of "In fact, as QuantumSpace noted, this is a general property". It's not like this is originally QuantumSpace's result (it is just a generic analysis proposition, after all), but I feel like it'd be nice to mention them since they posted the exact same result here on this page first. – Michael Jesurum Jun 17 '21 at 18:15
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    @MichaelJesurum, I don't know if Ramiro has read QuantumSpace answer or not when writing his own answer, but Ramiro's high knowledge has showed that most probably he doesn't need to read the first answer. –  Jun 17 '21 at 18:19
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    @MichaelJesurum , I had not seen QuantumSpace answer, when I wrote mine, BUT I agree with you. Thanks for the suggestion. I have updated my answer. – Ramiro Jun 17 '21 at 18:20
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    @LG I definitely believe that Ramiro already knew that result offhand, didn't mean to imply otherwise. – Michael Jesurum Jun 17 '21 at 18:24
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    Thanks a lot :) –  Jun 17 '21 at 19:52
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Claim: Let $f,g: X \to Y$ be two continuous functions (between metric spaces) such that $f$ and $g$ agree on a dense subspace $D$, then $f=g$.

Proof: Let $x \in X$ and choose a sequence $\{d_n\}_n$ in $D$ with $\lim_n d_n = x$. Then $$f(x)= \lim_n f(d_n) = \lim_n g(d_n) = g(x)$$ so $f=g$.

For your other question, the answer in that question proves the stronger statement that $L^1 \cap L^\infty$ is dense in $L^p$, so certainly $L^\infty$ is dense in $L^p.$

J. De Ro
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